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STRUCTURE OF POTASSIUM ISOTOPES
By Prof. Lefteris Kaliambos (Natural Philosopher in New Energy) (August 2014) Historically the discovery of the assumed uncharged neutron (1932) along with the invalid relativity (EXPERIMENTS REJECT RELATIVITY) led to the abandonment of the well-established electromagnetic laws, in favour of various contradicting nuclear theories, which could not lead to the nuclear structure. Under this physics crisis and using the charged UP and DOWN quarks , discovered by Gell-Mann and Zweig, I published my paper “Nuclear structure is governed by the fundamental laws of electromagnetism ” (2003), which led to my discovery of the new structure of protons and neutrons given by proton = + 5d + 4u = 288 quarks = mass of 1836.15 electrons neutron = + 4u + 8d = 288 quarks = mass of 1838.68 electrons The paper was also presented at a nuclear conference held at NCSR "Demokritos" (2002). In this photo I present the electromagnetic laws governing the nuclear structure, but a student of Einstein (Dr Th. Kalogeropoulos ) criticised my discovery of nuclear force and structure by believing that the nuclear structure is due to the invalid relativity. In fact, here one can see the 9 charged quarks in proton and the 12 ones in neutron able to give the charge distributions in nucleons for revealing the strong electromagnetic force for the nuclear binding in the correct nuclear structure by applying the laws of electromagnetism. You can see my papers of nuclear structure in my FUNDAMENTAL PHYSICS CONCEPTS . Note that according to my discovery of the LAW OF ENERGY AND MASS the mass defect in the nuclear structure is due to the photon mass of the emitting dipolic photon presented at the international conference "Frontiers of fundamental physics" (1993) organised by the natural philosophers M. Barone and F. Selleri , who gave me an award including a disc of the atomic philosopher Democritus. Nevertheless today many physicist continue to apply not the well-established laws but the various fallacious nuclear structure models which lead to complications . Potassium (K) has 25 known isotopes from K-32 to K-56. Three isotopes occur naturally: stable K-39 (93.3%) and K-41 (6.7%), and the long-lived radioisotope K-40 (0.012% Naturally occurring radioactive K-40 decays to stable Ar-40 (10.72% of decays) by electron capture or positron emission (giving it the longest known positron-emitter nuclide half-life). Alternately, and most of the time (89.28%), it decays to stable Ca-40 by beta decay. K-40 has a half-life of 1.248×109 years. The long half life of this primordial radioisotope is caused by a highly spin-forbidden transition: K-40 has a nuclear spin of -4, while both of its decay daughters are even-even isotopes with spins of 0. WHY K-39 AND K-41 WITH S = +3/2 ARE STABLE NUCLIDES ''' After a careful analysis of the structure of atomic nuclei I discovered that the beta decay is due to the fact that in unstable nuclei there exist single horizontal pn bonds of weak binding energy leading to the beta decay. For example in my paper STRUCTURE AND BINDING OF H3 AND He3 using the diagram of the structure of the H3 one sees that it is unstable because the two neutrons make single np bonds, while the He3 is stable because the one neutron between the two protons makes two np bonds per neutron. On the other hand the pp repulsions of long range lead to the instability when we have a small number of pn bonds per nucleon. For understanding the stable structure of K-39 you can read also my STRUCTURE OF Ar-40 AND K-39 . In the following diagram of K-39 you see that in the core of Mag-24 with S=0 the 6 deuterons like the n13p13, p14n14, n15p15, p16n16, n17p17 and p18n18 give S = +2, while the vertical pn system like the p19n19 gives S = 0. Since the extra n20 with two bonds per neutron at the second horizontal plane of negative spins (-HP2) gives S = -1/2 we get the total spin as S= 0 +2 +0 -1/2 = +3/2. '''DIAGRAM OF K-39 WITH S = +3/2 ' p12....n12 ' ' -HP6 n11......p11' ' p15......n10.....p10.......n18' ' +HP5 n15.......p9........n9.......p18 ' ' n14.......p8.......n8.......p17 ' ' -HP4 p14.......n7........p7.......n17 ' ' p13......n6........p6.......n16' ' +HP3 n13.......p5........n5.......p16 ' ' n19........p4........n4' ' -HP2 n3.......p3.......n20 ' ' p19........n2........p2' ' +HP1 p1........n1 ' Moreover in the presence of two extra neutrons of opposite spins with two bonds per neutron like the n21(+1/2) and the n22 (-1/2 we get the stable structure of K-41 with the same S = +3/2 as that of the K-39. That is, the stable structure of K-41 is based on the stable structure of K-39 because the two extra neutrons of opposite spins make two bonds per neutron. Especially the p1 and p19 form a blank position at +HP1 in order to receive the p21(+1/2). Whereas the p18 and p11 form a blank position at -HP6 in order to receive the p22(-1/2). ' ' NUCLEAR STRUCTURE OF THE UNSTABLE K-43, K-45, K-47 K-49, K-51, K-53 AND K-55 ' After a careful analysis I found that the structure of the above unstable nuclides is based on the structure of K-41 with S =+3/2 . For example the structure of K-43 with S =+3/2 has two more extra neutrons of opposite spins which form single bonds leading to the beta decay. ' ''' '''NUCLEAR STRUCTURE OF THE UNSTABLE K-37, K-35 AND K-33 WITH S = +3/2 Using the structure of K-39 with S =+3/2 we see that in the absence of two neutrons of opposite spins we get the structure of K-37. Also in the absence of 4 or 6 more neutrons we get the structures of K-35 and K-33 respectively. ' ' NUCLEAR STRUCTURE OF THE UNSTABLE K-40, K-42, K-44 K-46, K-48, K-50, K-52, AND K-54 In the following diagram of K-40 with S = -4 you see that the core is not the parallelepiped of Mg-24. In this case the structure is based on a parallelepiped with 5 horizontal planes like the -HP1, +HP2, -HP3 +HP4 and -HP5. Here the nucleons from p1 to n10 give a total S = -2. Also the deuterons from n1p1 to n15p15 give S=-1. Since the deuterons from n16p16 to p19n19 give S = 0 we see that for a structure of 19 protons and 19 neutrons we get a total spin S = -3. Note that in this structure the p2 and the p16 form a blank position at -HP1 in order to receive the p20(-1/2) with two bonds per neutron. Since there is not any more blank position the second extra p21(-1/2) at the -HP5 will make a single bond which leads to the beta decay . This structure has S = -4 given by S = -2 -1 +0 -1/2 -1/2 = -4. Then, after a careful analysis I found that the above unstable nuclides with extra neutrons of single bonds are based on the structure of K-40 with S = -4 and on the similar structure of 19 protons and 19 neutrons with S = -3. For example the K-42 with S = -2 is based on the structure of 19 protons and 19 neutrons with S = -3, in which we add two extra neutrons of positive spins and two extra neutrons of opposite spins. Also for getting the structure of K-54 with S = -2 we add 12 more extra neutron of opposite spins than those of the structure of K-42. DIAGRAM OF K-40 WITH S = - 4 ' n21.....p15.....n10......p10......n19' ' -HP5 n15.......p9.......n9........p19 ' ' n14.......p8........n8........p18 ' ' +HP4 p14......n7........ p7......n18 ' ' p13.......n6........p6........n17' ' -HP3 n13.......p5........n5.......p17 ' ' n12.......p4........n4........p16' ' +HP2 p12......n3........p3.......n16 ' ' p11......n2........p2.......n20' ' -HP1 n11.....p1.......n1 ' ' ' NUCLEAR STRUCTURE OF THE UNSTABLE K-38, K-36, K-34 AND K-32 Here the structure of the unstable K-38 with S =+3 is similar to the above structure of 19 protons and 19 neutrons which lead to the K-40 , because all nucleons change the spins . For example we have the +HP1, -HP2, +HP3, -HP4 and +HP5 giving S =+3. Then in the absence of two neutrons of positive spins we get the structure of K-36 with S =+2. That is S = +3 - 2(+1/2) = +2 . Then, in the absence of 2 more neutrons of positive spins we get the structure of K-34 with S = +1. That is S = +2 -2(+1/2) = +1. Finally in the absence of two more neutrons of opposite spins giving S=0 we get the structure of K-32 with the same S = +1. Category:Fundamental physics concepts